Question

Mrs. Bracken asks a random sample of 500 adults whether they favor giving vouchers to parents of school- age children that can be exchanged for education at any public or private school of their choice. Each school would be paid by the government on the basis of how many vouchers it collected. Suppose that 45%of the population favor this idea. If a second random sample of 500 is selected, what is the probability that the two sample proportions are less than 3 percentage points apart?"

Answer 1

Answer: The standard error of the sample proportion is given by:

SE = sqrt(p*(1-p)/n)

where p is the population proportion and n is the sample size.

For the first sample, the standard error is:

SE1 = sqrt(0.45*(1-0.45)/500) = 0.0284

For the second sample, the standard error is the same:

SE2 = sqrt(0.45*(1-0.45)/500) = 0.0284

The margin of error is given by:

ME = z*sqrt(SE1^2 + SE2^2)

where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, z = 1.96.

ME = 1.96*sqrt(0.0284^2 + 0.0284^2) = 0.0556

The two sample proportions can be less than 3 percentage points apart if the difference between them is less than 0.03:

|p1 - p2| < 0.03

The probability of this occurring can be calculated by finding the area under the normal distribution curve between -0.03 and 0.03, centered at the mean of the difference in sample proportions:

P(|p1 - p2| < 0.03) = P(-0.03 < p1 - p2 < 0.03)

mean = 0

standard deviation = sqrt(SE1^2 + SE2^2) = 0.0399

z = 0.03 / 0.0399 = 0.7519

Using a standard normal distribution table or calculator, we can find the probability to be approximately 0.229.

Therefore, the probability that the two sample proportions are less than 3 percentage points apart is approximately 0.229.

Explanation: